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Old 2019-09-08, 17:45   #1
paulunderwood
 
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Default A Flawed Proof of Fermat's Last Theorem

I just put the attached paper together. Comments will be welcomed.
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File Type: pdf FLT.pdf (43.2 KB, 60 views)

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Old 2019-09-08, 20:13   #2
pinhodecarlos
 
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Introduction: "Some 350 years ago", can't you be more accurate?

Introduction: Take out the double "famous"....this is not a contest who has the bigger..is the most famous mathematician (your opinion).


Where's the list of references?!


Edit: Need to go to bed, will read the rest on my return from A/L.

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Old 2019-09-08, 20:46   #3
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Quote:
Originally Posted by pinhodecarlos View Post
Introduction: "Some 350 years ago", can't you be more accurate?

Introduction: Take out the double "famous"....this is not a contest who has the bigger..is the most famous mathematician (your opinion).


Where's the list of references?!


Edit: Need to go to bed, will read the rest on my return from A/L.
I do not know who is the most famous mathematician and I don't think it is important, but, if we go by numbers of people who know of him I think it would be Isaac Newton.

I just wonder if this might have been the "proof" our friend Pierre found that it would not fit in the margin of the book Arithmetic by Diophantus? Now THAT was a famous mathematician.

But of course, all this is completely veering off the subject.
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Old 2019-09-08, 22:47   #4
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Thanks for your input. I have replaced the original document (see above) to include your suggestions and several corrected mistakes.

Last fiddled with by paulunderwood on 2019-09-08 at 22:49
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Old 2019-09-09, 08:11   #5
Nick
 
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I don't yet understand what you are trying to accomplish here.
(Once you assume something that is not actually true, anything you like can then be proved by contradiction, of course.)

Are you saying: "Fermat thought he had a proof and this may be what he had in mind"?
Is the assumption that F_n is prime for all n supposed to be the only flaw or is the paper supposed to be full of flaws?
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Old 2019-09-09, 09:14   #6
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Quote:
Originally Posted by Nick View Post
I don't yet understand what you are trying to accomplish here.
(Once you assume something that is not actually true, anything you like can then be proved by contradiction, of course.)

Are you saying: "Fermat thought he had a proof and this may be what he had in mind"?
Is the assumption that F_n is prime for all n supposed to be the only flaw or is the paper supposed to be full of flaws?
Yes, Fermat could have used the method(s) used in my paper. I do not have a time machine so that I could tell exactly.

I think F_n is prime is the only flaw, but I might be wrong!
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Old 2019-09-09, 18:12   #7
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I made a goof and so can simplify things greatly.

\(b^{2p}+c^{2p}\) can only be divisible by primes of the form \(4k+1\) (and maybe \(p\)). Hence there is no need to consider \(7,11,\ldots\) etc,

With this in mind, consider \(97 \mid U_p\) as an example. Then \(a^{8\cdot p \cdot 6} + b^{8\cdot p \cdot 6} \equiv 0 \pmod{97}\). But the unique "base" solution is \(9^2+4^2 \equiv 0 \pmod{97}\). Can I say \(9^4\) should be equivalent to unity? It is not, it is \(62\).

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Old 2019-09-09, 20:02   #8
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I am not claiming I have the full understatement of these issues.

But for instance, there are literally hundreds of conditional proofs assuming the RH (Riemann Hypothesis) is true.

A few of those proofs have been perfected and are no longer dependent on the RH to be true. In other words there are true independently of whether the RH is true or false.

(As an aside, this is considered some sort of validation of the RH being true. Not proof itself but evidence in favor.)

Now assume there is a conjecture lets call it the Steele, Richardson Conjecture that is considered valid if the RH is proven. Now furtherly assume that for some quirk or extraordinary breakthrough the RH is PROVEN FALSE.

Does this automatically mean The Steele-Richardson Conjecture is False?

I would contend that it does not. It only means that the conditional proof arrived at is no longer a valid path for proving the SR conjecture.

I do apologize for not using the exact mathematical language or terminology as I am just a dilettante, not a mathematician.

Last fiddled with by rudy235 on 2019-09-09 at 21:00 Reason: wether, weather, whether!
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Old 2019-09-10, 00:14   #9
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Is there any easy way to find the solutions to Fermat's prime as the sum of two squares other than by brute force trial and error?
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Old 2019-09-10, 01:24   #10
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Quote:
Originally Posted by rudy235 View Post
Does this automatically mean The Steele-Richardson Conjecture is False?

I would contend that it does not. It only means that the conditional proof arrived at is no longer a valid path for proving the SR conjecture.
You have the logic correct.

Consider the statement "if it snows, then it is cold outside."

On the 5th of March, it did not snow. Was it cold outside?

On the 10th of March, it was not cold outside. Did it snow?

So, a proof that goes "If RH is true, then XYZ" says nothing about XYZ when RH is false.

However, if you learn XYZ is false, then RH must be false.

Last fiddled with by VBCurtis on 2019-09-10 at 01:25
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Old 2019-09-10, 10:13   #11
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Quote:
Originally Posted by paulunderwood View Post
Is there any easy way to find the solutions to Fermat's prime as the sum of two squares other than by brute force trial and error?
This is equivalent to the problem of extending factorization algorithms from the integers to the Gaussian integers.
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